QQuestion
State and prove Pythagoras theorem.
Exam-ready answer • 05 Mark
✓Answer

Pythagoras Theorem: In any right-angled triangle, the area of the square drawn on the hypotenuse is equal to the sum of the areas of the squares drawn on the other two sides.
Given: ABC is a right-angled triangle whose ∠A is a right angle.
To prove: BC² = AB² + AC²
Construction: I draw a perpendicular AD on the hypotenuse BC from the right angular point A which intersects the side BC at the point D.
Proof: In right-angled Δ ABC, AD is perpendicular on the hypotenuse BC.
∴ Δ ABD and Δ CBA are similar.
Hence,
{\text{AB} \over \text{BC}} = {\text{BD} \over \text{AB}},
∴ AB² = BC × BD —- (i)
Again, ΔCAD and ΔCBA are similar.
Hence,
{\text{AC} \over \text{BC}} = {\text{DC} \over \text{AC}}
∴ AC² = BC × DC —- (ii)
So, by adding (i) and (ii), I get:
AB² + AC² = BC × BD + BC × DC.
= BC (BD + DC) = BC × BC = BC².
∴ BC² = AB² + AC² [proved]
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